Replace every array element by sum of previous and next
Last Updated :
14 Sep, 2022
Given an array of integers, update every element with sum of previous and next elements with following exceptions.
- First element is replaced by sum of first and second.
- Last element is replaced by sum of last and second last.
Examples:
Input : arr[] = { 2, 3, 4, 5, 6}
Output : 5 6 8 10 11
Explanation: We get the following array as {2+3, 2+4, 3+5, 4+6, 5+6}
Input : arr[] = { 3, 2, 1}
Output : 5 4 3
A Simple Solution is to create an auxiliary array, copy contents of given array to auxiliary array. Finally traverse the auxiliary array and update given array using copied values. Time complexity of this solution is O(n), but it requires O(n) extra space.
An efficient solution can solve the problem in O(n) time and O(1) space. The idea is to keep track of previous element in loop. Add the previous element using the extra variable and the next element to get each element.
Below is the implementation of this idea that is as follows:
C++
#include <iostream>
using namespace std;
void ReplaceElements( int arr[], int n)
{
if (n <= 1)
return ;
int prev = arr[0];
arr[0] = arr[0] + arr[1];
for ( int i = 1; i < n - 1; i++) {
int curr = arr[i];
arr[i] = prev + arr[i + 1];
prev = curr;
}
arr[n - 1] = prev + arr[n - 1];
}
int main()
{
int arr[] = { 2, 3, 4, 5, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
ReplaceElements(arr, n);
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
}
|
Java
import java.io.*;
class GFG {
static void ReplaceElements( int arr[], int n) {
if (n <= 1 ) {
return ;
}
int prev = arr[ 0 ];
arr[ 0 ] = arr[ 0 ] + arr[ 1 ];
for ( int i = 1 ; i < n - 1 ; i++) {
int curr = arr[i];
arr[i] = prev + arr[i + 1 ];
prev = curr;
}
arr[n - 1 ] = prev + arr[n - 1 ];
}
public static void main (String[] args) {
int arr[] = { 2 , 3 , 4 , 5 , 6 };
int n = arr.length;
ReplaceElements(arr, n);
for ( int i = 0 ; i < n; i++) {
System.out.print(arr[i] + " " );
}
}
}
|
Python 3
def ReplaceElements(arr, n):
if (n < = 1 ):
return
prev = arr[ 0 ]
arr[ 0 ] = arr[ 0 ] + arr[ 1 ]
for i in range ( 1 , n - 1 ):
curr = arr[i]
arr[i] = prev + arr[i + 1 ]
prev = curr
arr[n - 1 ] = prev + arr[n - 1 ]
if __name__ = = "__main__" :
arr = [ 2 , 3 , 4 , 5 , 6 ]
n = len (arr)
ReplaceElements(arr, n)
for i in range (n):
print (arr[i], end = " " )
|
C#
using System;
public class GFG {
static void ReplaceElements( int []arr, int n) {
if (n <= 1) {
return ;
}
int prev = arr[0];
arr[0] = arr[0] + arr[1];
for ( int i = 1; i < n - 1; i++) {
int curr = arr[i];
arr[i] = prev + arr[i + 1];
prev = curr;
}
arr[n - 1] = prev + arr[n - 1];
}
public static void Main() {
int []arr = {2, 3, 4, 5, 6};
int n = arr.Length;
ReplaceElements(arr, n);
for ( int i = 0; i < n; i++) {
Console.Write(arr[i] + " " );
}
}
}
|
PHP
<?php
function ReplaceElements( $arr , $n )
{
if ( $n <= 1)
return ;
$prev = $arr [0];
$arr [0] = $arr [0] + $arr [1];
for ( $i = 1; $i < $n - 1; $i ++)
{
$curr = $arr [ $i ];
$arr [ $i ] = $prev + $arr [ $i + 1];
$prev = $curr ;
}
$arr [ $n - 1] = $prev + $arr [ $n - 1];
return $arr ;
}
$arr = array (2, 3, 4, 5, 6);
$n = sizeof( $arr );
$arr1 = ReplaceElements( $arr , $n );
for ( $i = 0; $i < $n ; $i ++)
echo $arr1 [ $i ] . " " ;
?>
|
Javascript
<script>
function ReplaceElements(arr, n) {
if (n <= 1) {
return ;
}
let prev = arr[0];
arr[0] = arr[0] + arr[1];
for (let i = 1; i < n - 1; i++) {
let curr = arr[i];
arr[i] = prev + arr[i + 1];
prev = curr;
}
arr[n - 1] = prev + arr[n - 1];
}
let arr = [2, 3, 4, 5, 6];
let n = arr.length;
ReplaceElements(arr, n);
for (let i = 0; i < n; i++) {
document.write(arr[i] + " " );
}
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1) because it is using constant space for variables
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